Review of the Rham Cohomology Theory


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Let M be a smooth closed (i.e., compact and without boundary) manifold. Let TM and TM denote the tangent and cotangent (vector) bundle of M, respectively. Denote Λ(TM) the complex exterior algebra bundle of TM and Ω(M,C)=Γ(Λ(TM)) be the space of smooth sections of Λ(TM). Particularly, for any integer p with 0pn=dimM, we denote by
Ωp(M)=Γ(Λp(TM))


be the space of smooth p-forms over M.
 
Between the Ω(M,C), we have an operator d, i.e., the exterior differential operator, which maps a p-form to a p+1-form and with the property d2=0. Then, we obtain a complex chain (Ω(M,C),d)

Define the p-th (complex coefficient) de Rham cohomology of M as
HpdR(M,C)=kerd|Ωp(M,C)d(Ωp1(M,C)),

and the total de Rham cohomology of M as
HdR(M;C)=dimMp=0HpdR(M;C).

Particularly, a form ω is called closed if dω=0; called exact if there exists an η such that dη=ω.

Exercise 1. For general p the p-th de Rham cohomology may be hard to compute, but prove the following two special cases
  • H0dR(M;C) can be written as the direct sum of C with multiplicity be its connected components, i.e.,
    H0dR(M;C)=connected componentsC.

  • Suppose that M be connected, and define an functional
    M:Ωn(M;C)C,

    by ωMω, then prove that
    kerM=d(Ωn1(M;C)),

    from which we can compute the n-th de Rham cohomology HndR(M;C). (cf. [1])

Apparently HdR(M;C) is a vector space, since for any ω, ωΩ(M;C) (maybe one is k form and other is k form), then we can verify the following equation holds
[aω]=a[ω],[ω+ω]=[ω]+[ω],

where a is a constant function on M. With a litter more effort, we can prove that HdR(M;C) is a ring. In fact, define
[ω][ω]=[ωω],

then for any two differential forms η,η on M, we have (note that dω=0=dω)
(ω+dη)(ω+dη)=ωω+ωdη+dηω+dηdη=ωω+d((1)|ω|ωη+ηω+ηdη),

which means
[ω][ω]=[ωω].

Moreover, HdR(M;C) is superexchange, i.e.,
[ω][η]=(1)|ω||η|[η][ω].

When dimM=n=4m, consider H2mdR(M;C), then for any [ω], [η]H2mdR(M;C) we have
[ω][η]=[η][ω].


Define a bi-linear operator <,> on H2mdR(M;C) as
<[ω],[η]>=Mωη.

we can prove that <,> is a non-singular quadric on H2mdR(M;C) and its signature is a topological invariants, called the characteristic number of M.

Now, we state the de Rham theory without proof (for the proof cf. [1]).

Theorem 1. If M is a smooth closed orientable manifold, then for any integer p with 0pdimM, we have
  1. dimHpdR(M;C)<+;
  2. HpdR(M;C) is canonically isomorphic to Hpsing(M;C), the p-th singular cohomology of M.
Remark 1. The above Theorem shows that algebraic topology can be represented by differential forms, and Chern-Weil theory claims that the characteristic classes of vector bundle (which is a cohomology class) can be represented by de Rham cohomology class.

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